Mister Exam
Graphing y = ln(1+e^-x)
The solution
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/ -x\ f(x) = log\1 + E /
$$f{\left(x \right)} = \log{\left(1 + e^{- x} \right)}$$
f = log(1 + E^(-x))
The graph of the function
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
$$\log{\left(1 + e^{- x} \right)} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
so we need to solve the equation:
$$\log{\left(1 + e^{- x} \right)} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- \frac{e^{- x}}{1 + e^{- x}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- \frac{e^{- x}}{1 + e^{- x}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{\left(1 - \frac{e^{- x}}{1 + e^{- x}}\right) e^{- x}}{1 + e^{- x}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{\left(1 - \frac{e^{- x}}{1 + e^{- x}}\right) e^{- x}}{1 + e^{- x}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \log{\left(1 + e^{- x} \right)} = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \log{\left(1 + e^{- x} \right)} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
$$\lim_{x \to -\infty} \log{\left(1 + e^{- x} \right)} = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty} \log{\left(1 + e^{- x} \right)} = 0$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 0$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log(1 + E^(-x)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(1 + e^{- x} \right)}}{x}\right) = -1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = - x$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(1 + e^{- x} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
$$\lim_{x \to -\infty}\left(\frac{\log{\left(1 + e^{- x} \right)}}{x}\right) = -1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = - x$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(1 + e^{- x} \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
$$\log{\left(1 + e^{- x} \right)} = \log{\left(e^{x} + 1 \right)}$$
- No
$$\log{\left(1 + e^{- x} \right)} = - \log{\left(e^{x} + 1 \right)}$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\log{\left(1 + e^{- x} \right)} = \log{\left(e^{x} + 1 \right)}$$
- No
$$\log{\left(1 + e^{- x} \right)} = - \log{\left(e^{x} + 1 \right)}$$
- No
so, the function
not is
neither even, nor odd